A body of mass 2kg is placed on a horizontal frictionless surface. It is connected to one end of a spring whose force constant is 250N / m . The other end of the spring is joined with the wall. A particle of mass 0.15kg moving horizontally with speed v sticks to the body after collision. If it compresses the spring by 10cm , the velocity of the particle is

The correct option is 15 m/s.

Step 1: Understand the Collision (Conservation of Momentum)
The collision between the moving particle and the block is a completely inelastic collision because the particle sticks to the block after impact.
Let:
- Mass of the block, $M=2\text{kg}$
- Mass of the particle, $m=0.15\text{kg}$
- Initial velocity of the particle = $v$
- Common velocity of the combined mass immediately after collision = $V$

According to the law of conservation of linear momentum:

$mv=(M+m)V$
Substitute the given values:

$0.15v=(2+0.15)V$
$0.15v=2.15V$
$V=\frac{0.15}{2.15}v=\frac{3}{43}v$

Step 2: Conservation of Mechanical Energy
After the collision, the combined mass compresses the spring. The kinetic energy of the combined mass is completely converted into the elastic potential energy of the spring at maximum compression.
Let:
- Spring constant, $k=250\text{N/m}$
- Maximum compression, $x=10\text{cm}=0.1\text{m}$

Applying the conservation of mechanical energy:

$\frac{1}{2}(M+m)V^2=\frac{1}{2}k{x}^2$
$(M+m)V^2=k{x}^2$
Substitute the values:

$2.15\times V^2=250\times {\left(0.1\right)}^2$
$2.15V^2=250\times 0.01$
$2.15V^2=2.5$
$V^2=\frac{2.5}{2.15}\approx 1.163$
$V\approx 1.078\text{m/s}$

Step 3: Calculate the Initial Velocity of the Particle
Substitute the value of $V$ back into the momentum equation:

$1.078=\frac{3}{43}v$
$v=1.078\times \frac{43}{3}$
$v\approx 15.45\text{m/s}$

Rounding to the nearest integer value given in the options, we get:

$v\approx 15\text{m/s}$